I have this problem about finding the mixed Nash equilibrium. The payoff matrix is the following
A(p) B(q) C(1-p-q)
A 4 0 0
B 0 4 0
C 3 3 2
The only method that I know is to compute the expectation value of stategies of one player with respect to the other. So I obtain:
E(A)=4p
E(B)=4q
E(C)=3p+3q+2-2p-2q
and now I impose E(A)=E(B) and E(B)=E(C) and from the first equation I get p=q and from the second one p=1.. But this is not correct because the sum oh the probabilities should be one, so I have problem with equation p=q. I thought that this procedure for finding mixed Nash equilibrium was correct, but it seems not. So, where is my mistake? Thank you in advance!
If this is a zero-sum game, as your question suggests, then there is a unique pure-strategy equilibrium. If the column player plays the rightmost column, the unique best response of the row player is the bottom row. Against the bottom row, the best response of the column player is the rightmost column (the column player is minimizing).
In general, you are right that if a player is to play multiple pure strategies in equilibrium then they must be indifferent between the strategies they play. But actually that's not enough for a mixed strategy to be a best response (since there could be other pure strategies that give higher payoff). See these answers for more details:
Finding mixed nash equlibrium
Calculating Nash equilibrium in mixed strategy in a game where a Nash equilibrium in pure strategy exists
You should also check out these web-based game solvers:
http://banach.lse.ac.uk/
Game Theory Explorer: http://gte.csc.liv.ac.uk/gte/builder/